A polynomial with integer coefficients is of the form
\[28x^4 + a_3 x^3 + a_2 x^2 + a_1 x + 1 = 360.\]
Find the number of different possible rational roots of this polynomial.
Adding $-360$ to both sides gives \[28x^4 + a_3 x^3 + a_2 x^2 + a_1 x - 359 = 0,\] and so by the Rational Root Theorem, its rational roots are of the form $\pm \frac{p}{q},$ where $p$ divides $-359$ and $q$ divides 28. Since $359 = 359 \cdot 1,$ the only possibilities for $p$ are $\pm 1$ and $\pm 359.$ Since $28 = 2^2 \cdot 7,$ the only possibilities for $q$ are $\pm 1, \pm 2, \pm 4, \pm 7, \pm 14,$ and $\pm 28.$ The possible values for $x$ are then
\begin{align*}
x &= \frac{1}{1}, -\frac{1}{1}, \frac{359}{1}, -\frac{359}{1} \quad \text{and} \\
x &= \frac{1}{2}, -\frac{1}{2}, \frac{359}{2}, -\frac{359}{2}\\
x &= \frac{1}{4}, -\frac{1}{4}, \frac{359}{4}, -\frac{359}{4}\\
x &= \frac{1}{7}, -\frac{1}{7}, \frac{359}{7}, -\frac{359}{7}\\
x &= \frac{1}{14}, -\frac{1}{14}, \frac{359}{14}, -\frac{359}{14}\\
x &= \frac{1}{28}, -\frac{1}{28}, \frac{359}{28}, -\frac{359}{28}.
\end{align*}This is a total of $2 + 2 + 2 + 2 + 2 + 2 = \boxed{12}$ values for $x.$
To find the number of different possible rational roots of the polynomial, we can use the Rational Root Theorem. The Rational Root Theorem states that if a polynomial with integer coefficients has a rational root \(\frac{p}{q}\), where \(p\) and \(q\) are integers, then \(p\) must be a factor of the constant term (in this case, 1) and \(q\) must be a factor of the leading coefficient (in this case, 28).
So, we need to find the factors of 1 and 28.
The factors of 1 are \(\pm1\).
The factors of 28 are \(\pm1\), \(\pm2\), \(\pm4\), \(\pm7\), \(\pm14\), and \(\pm28\).
Combining these factors, the possible rational roots are:
\[\pm1, \pm2, \pm4, \pm7, \pm14, \pm28.\]
Hence, there are 12 different possible rational roots for this polynomial.